Mathematics, 30.01.2020 04:43 sonaihriley

If matrix a has dimensions m x n and matrix b has dimensions n x p where m, n, and p are distinct positive integers, which of the following is true?
i. the product of ba doesn't exist
ii. the product of ab exists and has dimensions m x p
iii. the product of ab exists and has dimensions n x n

a) i only
b) ii only
c) iii only
d) i and ii only
e) i and iii

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You probably already have an idea of what a matrix is; it's a rectangular array of numbers. What they represent is a bit complicated to explain. There's a whole subject about it (see "linear algebra" for more info). Whatever they represent isn't important though, you don't need to know everything about matrices to compute their product (or whether it's even possible).

A quick definition: A matrix of dimension -by- is a matrix with rows and columns.

Matrix multiplication all comes down to an operation called the "dot product". It's defined by the sum of component-wise products of elements between two lists. What this means is, if $x=\{1,2,0\}$ and $y=\{-1,0,3\}$ , then the dot product of and is

$x\cdot y=(1)(-1)+(2)(0)+(0)(3)=-1+0+0=-1$

What we did was take the first elements of each list and multiplied them, and the same for the other two elements, then added them all together. Notice that the product can't be computed if and don't thave the same number of elements.

We write this product in matrix form as

$\begin{bmatrix}1&2&0\end{bmatrix}\begin{bmatrix}-1\\0\\3\end{bmatrix}$

Notice the pattern here: on the left, a matrix with 1 row and 3 columns; on the right, a matrix with 3 rows and 1 column. The number of columns of the first matrix have to match the number of rows of the second.

The orientation makes a big difference. The product above returns a 1-by-1 matrix (or simply a scalar number):

$\begin{bmatrix}1&2&0\end{bmatrix}\begin{bmatrix}-1\\0\\3\end{bmatrix}=\begin{bmatrix}(1)(-1)+(2)(0)+(0)(3)\end{bmatrix}=\begin{bmatrix}-1\end{bmatrix}=-1$

On the other hand, the alternate orientation would result in a 3-by-3 matrix.

$\begin{bmatrix}1\\2\\0\end{bmatrix}\begin{bmatrix}-1&0&3\end{bmatrix}=\begin{bmatrix}(1)(-1)&(1)(0)&(1)(3)\\(2)(-1)&(2)(0)&(2)(3)\\(0)(-1)&(0)(0)&(0)(3)\end{bmatrix}=\begin{bmatrix}-1&0&3\\\-2&0&3\\0&0&0\end{bmatrix}$

So the number of rows of the first matrix and number of columns of the second matrix determine the number of rows and columns, respectively, of the matrix product.

I think we have enough information about matrix multiplication to answer this question. If has dimensions $m\times n$ and has dimensions $n\times p$ , then the matrix product exists ( columns in , rows in ), but the matrix product does not ( columns in , rows in , but $m\neq p$ ). So I is not true.

We know exists, and with rows in and columns in , we expect to have rows and columns, so has dimensions $m\times p$ . So II is true.

With dimensions $r\times c$ , a matrix would contain elements. m,n,p are distinct, so $mp\neq n^2$ . So III is not true.

Answer from: Quest

y = - 6

step-by-step explanation:

given the point (- 2, y) lies on the graph then it is a solution of it's equation.

substitute x = - 2 into the equation for value of y

y = 3(- 2) = 3 × - 2 = - 6

the point on the graph is (- 2, 6)

Answer from: Quest

The second step on the test was the move he was supposed to

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If matrix a has dimensions m x n and matrix b has dimensions n x p where m, n, and p are distinct po...